Population Genetics Unit 4 AP Biology Population Genetics
Population Genetics Unit 4 AP Biology
Population Genetics • Study of the frequency of particular alleles and genotypes in a population • Ex. Suppose you want to determine the % of alleles and genotypes for Alu in the PV 92 region.
Sample Data. Alu in the PV 92 region • 100 AP Biology students total 45 students are +/+ 20 students are +/35 students are -/-
Allelic Frequencies & Genotype Frequencies • Genotype Frequency – % of a particular genotype that is present in a population • Allelic Frequency – % of particular allele in a population
Sample Problem • 100 AP Biology students total – 45 students are +/+ – 20 students are +/– 35 students are -/- • Calculate the genotype frequencies • Calculate the allelic frequencies
Solution- Genotype Frequencies • Genotype frequency of +/+ 45 (+/+) / 100 students total = 0. 45 • Genotype Frequency of +/20 (+/-) / 100 students total = 0. 20 • Genotype Frequency of -/– 35 (-/-) / 100 students total = 0. 35
Solution- Allelic Frequencies • Step 1: Calculate the total # of alleles • 100 students x 2 alleles each = 200 total alleles
Solution- Allelic Frequencies • Allelic Frequency of “+” allele – Each student with +/+ contributes 2 “+” alleles each 2 x 45 = 90 “+” alleles – Each student with +/- contributes 1 “+” allele each 1 x 20 = 20 “+” alleles – 90 + 20 = 110 “+” alleles – Allelic Frequency = 110 “+” alleles /200 total = 0. 55
Solution- Allelic Frequencies • Allelic Frequency of “-” allele – Each student with -/- contributes 2 “-” alleles each 2 x 35 = 70 “-” alleles – Each student with +/- contributes 1 “-” allele each 1 x 20 = 20 “-” alleles – 70 + 20 = 90 “-” alleles – Allelic Frequency = 90 “-” alleles /200 total = 0. 45
Practice Problem #1 • 325 ants total: – 140 ants have the genotype GG – 75 ants have the genotype Gg – 110 ants have the genotype gg • Calculate the genotype and allelic frequencies.
Solution #1 • Genotype Frequencies: – GG = 140 / 325 = 0. 43 – Gg = 75 / 325 = 0. 23 – gg = 110 / 325 = 0. 34
Solution #1 • How many alleles total? 325 X 2 = 650 alleles total • Allelic Frequencies: – G : ( (140 x 2) + 75) / 650 = 0. 55 – g : ( (110 x 2) + 75 ) / 650 = 0. 45
Question… • What should all of the allelic frequencies add up to? – They should add up to 1 • What should all of the genotype frequencies add up to? – They should add up to 1
Gene Pool • All the alleles in a population • Alleles in the gene pool can change as a result of several different factors (mutations, immigration, natural selection)
Evolution • If the genotype and allelic frequencies are NOT changing from generation to generation the population is NOT evolving.
Hardy Weinberg Equilibrium • Mathematical model to describe a nonevolving population • In real life, populations are almost never in Hardy Weinberg Equilibrium • 2 variables: p = allelic frequency of one allele q = allelic frequency of other allele
Conditions of Hardy Weinberg • • • Very large population size No Gene Flow (no moving in or out) No Mutations (no new alleles introduced) No Sexual Selection No Natural Selection (no allele causes the individual to survive better or worse) • Not meeting these conditions would cause the allelic and genotype frequencies to change
Hardy Weinberg Equations • p and q are frequencies of alleles in a population • p + q = 1 (the 2 alleles make up 100% of the alleles) • From this, there are 3 genotypes possible: – Homozygous dominant (Ex. HH) – Homozygous recessive (Ex. hh) – Heterozygous (Ex. Hh)
Hardy Weinberg Equations • Using the allelic frequencies (p and q), genotype frequencies can be calculated for each genotype • p 2 = genotype frequency for homozygous dominant (HH) • 2 pq = genotype frequency for heterozygous genotype (Hh) • q 2 = genotype frequency for homozygous recessive (hh)
Hardy Weinberg Equations • Equation: p 2 + 2 pq + q 2 = 1 • Why is it 2 pq? – Because there are two heterozygous combinations possible (Hh and h. H)
Hardy Weinberg Equations • Based on probability • p = 0. 8 (allelic frequency for CR) • q = 0. 2 (allelic frequency for CW) • Chance of CRCR = 0. 8 x 0. 8 = 0. 64
Population in HW Equilibrium? • You can use calculations to determine if a population is in Hardy Weinberg Equilibrium • Let’s take practice problem #1: – Actual allelic frequencies are: • p = 0. 55 q = 0. 45 – Actual genotype frequencies are: • 0. 43 (GG), 0. 23 (Gg), 0. 34 (gg)
If the population is in HW Equilibrium… • If the population is in Hardy Weinberg equilibrium, then the calculated expected genotype frequencies should match the actual genotype frequencies. • Using the Hardy Weinberg Equation: – Expected GG Genotype frequency = p 2 – Expected Gg Genotype frequency = 2 pq – Expected gg genotype frequency = q 2
Expected Genotype Frequencies • p = 0. 55, q = 0. 45 • Expected Genotype frequencies: – GG = p 2 = (0. 55)2 = 0. 30 – Gg = 2 pq = 2 (0. 55)(0. 45) = 0. 50 – gg = q 2 = (0. 45)2 = 0. 20 – Actual genotype frequencies are: • 0. 43 (GG), 0. 23 (Gg), 0. 34 (gg) – The actual genotype and expected genotype frequencies do not match the population is NOT in HW Equilibrium.
One more Practice Problem • You are studying a ferret population for their nose sizes. B is the allele for a long nose, b is the allele for a short nose. • 78 ferrets are BB • 65 ferrets are Bb • 21 ferrets are bb • What are the actual allelic and genotype frequencies? Is the population in HW Equilibrium?
Solution • Actual genotype frequencies: – BB = 78 / 164 = 0. 47 – Bb = 65 / 164 = 0. 40 – bb = 21 / 164 = 0. 13
Solution • Actual Allelic frequencies: – Total alleles = 2 x 164 = 328 – Allelic frequency of B = p = ( (2 x 78) + 65) /328 = 0. 67 – Allelic frequency of b = q = ( (2 x 21) + 65) / 328 = 0. 33
Solution • Expected genotype frequencies: – BB = p 2 = (0. 67)2 = 0. 45 – Bb = 2 pq = 2(0. 67)(0. 33) = 0. 44 – bb = q 2 = (0. 33)2 = 0. 11 • These do not match the actual frequencies, so the population does not appear to be in HW Equilibrium.
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